1015 Reversible Primes (20 point(s))

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本文介绍了一种算法，用于判断一个给定的正整数N是否为D进制下的可逆素数，即N及其在D进制下的反转数均为素数。文章提供了输入输出规范、样例和代码实现，包括数字反转和素数判断两个关键函数。

1015 Reversible Primes (20 point(s))

A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (<105) and D (1<D≤10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line Yes if N is a reversible prime with radix D, or No if not.

Sample Input:

Sample Output:

Yes
Yes
No

Tips:

（1）在D进制下把数字n反转：想一想是不是和在10进制下反转是一样的呢？并不用先转换为D进制，再反转，再转换为10进制。

（2）此题素数判断要求到了1e5，使用素数筛法会造成MLE。

#include<iostream>
using namespace std;
int reverse(int n,int d){//reverse any number with d redix
	int ans = 0;
	while(n){
		ans = ans*d+n%d;
		n/=d;
	}
	return ans;
}
bool isPrime(int x){//judge if it is a prime number
	if(x<2) return false;
	for(int i=2;i*i<=x;i++){
		if(x%i==0) return false;
	}
	return true;
}
int main(void){
	int n,d;
	while(cin>>n){
		if(n<0) break;
		cin>>d;
		if(isPrime(n)&&isPrime(reverse(n,d)))cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}
	return 0;
}